# How do you differentiate p = 2log_3(5^s) - log_3(4^s)?

Jan 24, 2017

$\frac{\mathrm{dp}}{\mathrm{ds}} = {\log}_{3} \left(\frac{25}{4}\right)$

#### Explanation:

Note we can use the rule ${\log}_{a} \left({b}^{c}\right) = c {\log}_{a} \left(b\right)$. Then:

$p = 2 s {\log}_{3} \left(5\right) - s {\log}_{3} \left(4\right)$

Note that $2 {\log}_{3} \left(5\right)$ and ${\log}_{3} \left(4\right)$ are both constants, so the derivative is simply:

$\frac{\mathrm{dp}}{\mathrm{ds}} = 2 {\log}_{3} \left(5\right) - {\log}_{3} \left(4\right)$

If you wish, use $c {\log}_{a} \left(b\right) = {\log}_{a} \left({b}^{c}\right)$ and ${\log}_{m} \left(n\right) - {\log}_{m} \left(q\right) = {\log}_{m} \left(\frac{n}{q}\right)$.

$\frac{\mathrm{dp}}{\mathrm{ds}} = {\log}_{3} \left(\frac{25}{4}\right)$